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We have 2 arrays of objects of the same class. How to substract the 1st from the 2nd, based only on one property?

AxelleDRouge on December 08, 2019

So this is the first technical post I am writing. It is mostly destined to my future self, but also, I am not really satisfied with the answer to t...

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Ankit kumar shukla • Dec 8 '19

Merge both aaray using
Let newSet=[...set1,...set2]
Then apply this funtiin...
function unique(array, propertyName) {
return array.filter((e, i) => array.findIndex(a => a[propertyName] === e[propertyName]) === i);
}

AxelleDRouge • Dec 8 '19 • Edited

Thanks for your reply.
Could you details more the process you use? Why would you use findIndex?
I tried to use your method in my use case, but it doesn't remove all the values from the smaller list in the total one
I will edit my post to make it more clearer with the expected result

Ankit kumar shukla • Dec 8 '19

I have used findindex to find current index property value to match with passed property to make it unique

AxelleDRouge • Dec 8 '19

Doesn't findIndex returns the first value that match the provided testing function? In need all the elements that match the predicate, not only the first one.

Webdeveloper⭐⭐ • May 18 '20

a more concise way would be to use the new ES6 spread operator and Set
merge your 2 arrays then
let mergedArray = array1.concat(array2)
const unique = [...new Set(mergedArray.map(item => item.label))];

Dhilip kumar • Dec 8 '19

I'm no expert just tried.

Not tested.

Store all the various props of arrayToSubtract in an Object.

var referenceObj = {}
arrayToSubstract.forEach((item) => referenceObj[item[propertyOfReference]] = true;);

//Reference object keys should not be in arrayTotal

var outputArr = arrayTotal.filter((item) => !referenceObj[item[propertyOfReference]);

AxelleDRouge • Dec 8 '19

Interesting way to resolve the problem, it should work :)

Duy K. Bui • Dec 8 '19

First of all:

Why would you define functions within your function and call them each only once? I would just skip all those arrow functions.
Use const in place of let unless you absolutely need to reassign these variables, which you don't in this case.

Your solution seems to work, only 2 things I can think of to improve (besides not defining functions then calling them once as mentioned above):

You are calling extractValue() on the SAME set of parameters for every member of arrayTotal, that's redundant.
You can replace that .every() call with .includes().

A revised version would look like this:

const substractObjectsFromArray = (arrayTotal, arrayToSubstract, propertyOfReference) => {
  const idsToSubtract = arrayToSubtract.map(element => element[propertyOfReference]);
  return arrayTotal.filter(element => idsToSubtract.includes(element[propertyOfReference]);
}

AxelleDRouge • Dec 8 '19

Thanks you are right, it's a good way to improve it.